Problem: Find $\lim_{x\to 1^+}(2x-1)^{^{\frac{1}{x-1}}}$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $e^{-2}$ (Choice C) C $e^2$ (Choice D) D The limit doesn't exist.
Substituting $x=1$ into $(2x-1)^{^{\frac{1}{x-1}}}$ results in the indeterminate form $1^{^{\infty}}$. To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting $y=(2x-1)^{^{\frac{1}{x-1}}}$, we will find $\lim_{x\to 0}\ln(y)$. Once we find it, we will be able to find $\lim_{x\to 1^+}y$. $\ln(y) =\dfrac{\ln(2x-1)}{x-1}$ Substituting $x=1$ into $\dfrac{\ln(2x-1)}{x-1}$ results in the indeterminate form $\dfrac{0}{0}$, so now it's L'Hôpital's rule's turn to help us with our quest! $\begin{aligned} &\phantom{=}\lim_{x\to 1^+}\ln(y) \\\\ &=\lim_{x\to 1^+}\dfrac{\ln(2x-1)}{x-1} \\\\ &=\lim_{x\to 1^+}\dfrac{\dfrac{d}{dx}[\ln(2x-1)]}{\dfrac{d}{dx}[x-1]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 1^+}\dfrac{\left(\dfrac{2}{2x-1}\right)}{1} \\\\ &=\dfrac{\left(\dfrac{2}{1}\right)}{1} \gray{\text{Substitution}} \\\\ &=2 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 1^+}\dfrac{\dfrac{d}{dx}[\ln(2x-1)]}{\dfrac{d}{dx}[x-1]}$ actually exists. We found that $\lim_{x\to 1^+}\ln(y)=2$, which means $\lim_{x\to 1^+}y=e^2$. [Why?] In conclusion, $\lim_{x\to 1^+}(2x-1)^{^{\frac{1}{x-1}}}=e^2$.